Letter E is correct.
Area of equilateral trianlge= (1/2) tiime a^2 square root 3
Equation for the base is = x^2+y^2 =9
at x ,
y= square root (9-x^2)
Side of triangle a= 2y
d times Volume = (1/2)a^2 square root 3 dx
d times V= (square root 3) (9-x^2) dx
V= square root 3 INT (9-x^2) dx
-3V= 2square root INT (9-x^2) dx
0
V= 2 square root 3 ( 9x-x^3/3)
V=2 square root 3 ( 27-9)
V= 36 square root 3
The answer is E.
Answer:
The largest total area that can be enclosed will be a square of length 272 yards.
Step-by-step explanation:
First we get the perimeter of the large rectangular enclosure.
Perimeter of a rectangle =2(l + w)
Perimeter of the large rectangular enclosure= 1088 yard
Therefore:
2(L+W)=1088
The region inside the fence is the area
Area: A = LW
We need to solve the perimeter formula for either the length or width.
2L+ 2W= 1088 yd
2W= 1088– 2L
W = 
W = 544–L
Now substitute W = 544–L into the area formula
A = LW
A = L(544 – L)
A = 544L–L²
Since A is a quadratic expression, we re-write the expression with the exponents in descending order.
A = –L²+544L
Next, we look for the value of the x coordinate


L=272 yards
Plugging L=272 yards into the calculation for area:
A = –L²+544L
A(272)=-272²+544(272)
=73984 square yards
Thus the largest area that could be encompassed would be a square where each side has a length of 272 yards and a width of:
W = 544 – L
= 544 – 272
= 272 yards
V=1/3Bh B=area of base h=height
Step-by-step explanation:
he doesnt round off his value,,,he leaves divided the values wrongly ,,he never left his work in 4 significant figures
Answer:
answer is 6
Step-by-step explanation:
48/8=6
hope this helps!